Problem: Simplify the following expression: $\dfrac{12n^2}{21n^3}$ You can assume $n \neq 0$.
Answer: $ \dfrac{12n^2}{21n^3} = \dfrac{12}{21} \cdot \dfrac{n^2}{n^3} $ To simplify $\frac{12}{21}$ , find the greatest common factor (GCD) of $12$ and $21$ $12 = 2 \cdot 2 \cdot 3$ $21 = 3 \cdot 7$ $ \mbox{GCD}(12, 21) = 3 $ $ \dfrac{12}{21} \cdot \dfrac{n^2}{n^3} = \dfrac{3 \cdot 4}{3 \cdot 7} \cdot \dfrac{n^2}{n^3} $ $\phantom{ \dfrac{12}{21} \cdot \dfrac{2}{3}} = \dfrac{4}{7} \cdot \dfrac{n^2}{n^3} $ $ \dfrac{n^2}{n^3} = \dfrac{n \cdot n}{n \cdot n \cdot n} = \dfrac{1}{n} $ $ \dfrac{4}{7} \cdot \dfrac{1}{n} = \dfrac{4}{7n} $